Mathematical-Reasoning · Unit 10 · Video 1 · Interactive Practice
Many Ways to Write 27 = 4q + r, and a Mod 6 Proof
IKey Formulas
Statement
Name
What it says
a=bq+r, so r=a−bq
Writing a with b
Every integer q gives exactly one integer r; raising q by 1 lowers r by b
27=4⋅6+3, −17=5(−4)+3
Quotient and remainder
The remainder is the smallest nonnegative r (zero allowed); the quotient is the q that goes with it
a≡b(modn)⟺n∣(a−b)
Congruence modulo n
For n∈N: a−b=nk, that is, a=b+nk, for some k∈Z
(a+b)−2=6(k+m+1)
The mod 6 proof
If a≡3 and b≡5(mod6), then a+b≡2(mod6)
Key Insight: Many pairs (q,r) satisfy a=bq+r; insisting that r be the smallest nonnegative one picks out a single pair. A congruence proof runs from congruence to divides, from divides to an equation with an integer that must exist, then algebra, and finally factoring out the modulus.
IIOne Equation, Many Pairs, One Remainder
Every choice of q gives one r=a−bq; the remainder is the smallest of these that is not negative.
IIISums from the Sets A and B, Modulo 6
For a≡3(mod6) and b≡5(mod6), every sum a+b lands 2 past a multiple of 6.
💡 Any number of checked pairs is evidence, not proof: the conjecture is about infinitely many pairs of integers, so only an argument from the definitions can settle it.
IVWhere the 2 Comes From
In (3+6k)+(5+6m), the leftover 3+5=8 is one more 6 plus 2.
Step 1 — Unpack the hypothesesa−3=6k,b−5=6m,k,m∈Za=3+6k,b=5+6m
Step 2 — Add the two equationsa+b=(3+6k)+(5+6m)=8+6k+6m
Step 3 — Subtract 2 from both sides(a+b)−2=6+6k+6m
Step 4 — Factor out the modulus(a+b)−2=6(k+m+1),k+m+1∈Z
The integers are closed under addition, so 6∣(a+b)−2, and therefore a+b≡2(mod6). ■
VQuiz Questions
Problem 1 · Which Pair Is the Remainder?
Given: all four equations below are true. Which one shows the quotient and remainder when 38 is divided by 7, using the convention that the remainder is the smallest nonnegative r with 38=7q+r?
✅ Correct!r=3 is nonnegative, and one more 7 would make it 3−7=−4, which is negative. So the quotient is 5 and the remainder is 3.
❌ Not quite. Here r=−4 satisfies the equation, but it is negative, so the convention rules it out.
❌ Not quite.r=10 is nonnegative, but it still contains a 7: taking q=5 leaves r=3, which is smaller and still nonnegative.
❌ Not quite.r=17 still contains two more 7s; raising q lowers r by 7 each time, down to r=3 at q=5.
Show solution
Solve 38=7q+r for r: r=38−7q. Raising q by 1 lowers r by 7:
q=3,4,5,6⟶r=17,10,3,−4
The nonnegative values are 17, 10 and 3, and the smallest is 3, at q=5. So 38=7⋅5+3: the quotient is 5 and the remainder is 3, as long division gives.
Problem 2 · Dividing a Negative Integer
Given:−23 is divided by 6. Using the convention that the remainder is the smallest nonnegative r with −23=6q+r, find the quotient q and remainder r.
✅ Correct!6(−4)+1=−24+1=−23, and r=1 is the smallest nonnegative choice.
❌ Close, but…6(−3)+(−5)=−23 is true, but r=−5 is negative. The convention requires a nonnegative remainder.
❌ Not quite. Check the equation: 6(−3)+5=−18+5=−13, not −23. The remainder of −23 is not the remainder of 23 with a sign attached.
❌ Not quite.6(−5)+7=−23 is true and r=7 is nonnegative, but it is not the smallest: q=−4 leaves r=1.
Show solution
From −23=6q+r we get r=−23−6q. Each step up in q lowers r by 6:
q=−6,−5,−4,−3⟶r=13,7,1,−5
The smallest nonnegative value is r=1, at q=−4. Hence
−23=6(−4)+1,
so the quotient is −4 and the remainder is 1. The pair q=−3, r=−5 also satisfies the equation, but −5 is negative, so it is ruled out.
Problem 3 · The Same Proof, Modulo 7
Given: integers a and b with a≡4(mod7) and b≡6(mod7), written as a=4+7k and b=6+7m with k,m∈Z.
Which integer r with 0≤r<7 satisfies a+b≡r(mod7) for every such a and b?
Which integer t makes (a+b)−r=7t?
✅ Correct!(a+b)−3=7+7k+7m=7(k+m+1), and k+m+1 is an integer, so a+b≡3(mod7).
❌ Not quite.a+b≡10(mod7) is true, but 10 is not less than 7: 10=7+3.
❌ Not quite.2 was the answer for the mod 6 example. Here the leftover is 4+6=10=7+3.
❌ Not quite.10−6=4 removes a 6, the video's modulus; here the modulus is 7, so remove a 7.
❌ Not quite. Add the leftovers 4+6 and remove as many 7s as possible.
❌ Not quite.7(k+m)=7k+7m, but (a+b)−3=7+7k+7m; the extra 7 from 10=7+3 is missing.
❌ Not quite.7(k+m+3)=21+7k+7m, which is 14 too big. Only one extra 7 comes out of 10−3.
❌ Not quite.7(10+k+m)=70+7k+7m. The 10 is the leftover before subtracting r, and it is not multiplied by 7.
❌ Not quite. Expand (a+b)−3 and factor out 7.
Show solution
By the definition of congruence, 7∣(a−4) and 7∣(b−6), so a=4+7k and b=6+7m for some integers k and m. Adding,
a+b=10+7k+7m.
Since 10=7+3, subtracting 3 from both sides gives
(a+b)−3=7+7k+7m=7(k+m+1).
Because k+m+1∈Z, 7∣(a+b)−3, so a+b≡3(mod7), with r=3 and t=k+m+1. Check: a=4, b=6 gives 10−3=7=7⋅1, matching t=0+0+1.
Problem 4 · Spot the Flaw in a Proof
Given: a student proving "if a≡3(mod6) and b≡5(mod6), then a+b≡2(mod6)" writes: "There is an integer k with a=3+6k and b=5+6k. Then (a+b)−2=6+12k=6(2k+1), so 6∣(a+b)−2." What is wrong?
✅ Correct! The two hypotheses each supply their own integer: a−3=6k and b−5=6m. Reusing k silently adds the assumption k=m, which forces b=a+2.
❌ Not quite. Once a is fixed, k=(a−3)/6 is fixed too. With the same k, b must equal a+2; for a=15, b=11 we have k=2 but (11−5)/6=1.
❌ Not quite. That still ties b to a (it forces b=a+8). The integer for b must be a new, independent letter.
❌ Not quite. Examples are evidence, never part of a proof. The flaw is in how the hypotheses were translated into equations.
Show solution
From a≡3(mod6) the definition gives some integer k with a−3=6k. From b≡5(mod6) it gives some integer with b−5=6⋅(that integer), and nothing says it equals k. Writing b=5+6k assumes b−a=2, so the argument proves the statement only for pairs like a=9, b=11.